# ch8 - 8.1.IDENTIFY and SET UP EXECUTE(b(i v = p = mv K = 1...

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8.1.I DENTIFY and S ET U P : . p mv = 2 1 2 . K mv = E XECUTE : (a) 5 (10,000 kg)(12.0 m/s) 1.20 10 kg m/s p = = × (b) (i) 5 1.20 10 kg m/s 60.0 m/s 2000 kg p v m × = = = . (ii) 2 2 1 1 T T SUV SUV 2 2 m v m v = , so T SUV T SUV 10,000 kg (12.0 m/s) 26.8 m/s 2000 kg m v v m = = = E VALUATE : The SUV must have less speed to have the same kinetic energy as the truck than to have the same momentum as the truck. 8.7.I DENTIFY : The average force on an object and the object’s change in momentum are related by Eq. 8.9. The weight of the ball is . w mg = S ET U P : Let + x be in the direction of the final velocity of the ball, so 1 0 x v = and . 2 25.0 m/s x v = E XECUTE : av 2 1 2 1 ( ) ( ) x x F t t mv mv = x gives 2 1 av 3 2 1 (0.0450 kg)(25.0 m/s) ( ) 562 N 2.00 10 s x x x mv mv F t t = = = × . . The force exerted by the club is much greater than the weight of the ball, so the effect of the weight of the ball during the time of contact is not significant. 2 (0.0450 kg)(9.80 m/s ) 0.441 N w = = E VALUATE : Forces exerted during collisions typically are very large but act for a short time. 8.9.I DENTIFY : Use Eq. 8.9. We know the intial momentum and the impluse so can solve for the final momentum and then the final velocity. S ET U P : Take the x -axis to be toward the right, so Use Eq. 8.5 to calculate the impulse, since the force is constant. 1 3.00 m/s. x v = + E XECUTE : (a) 2 1 x x x J p p = 2 1 ( ) ( 25.0 N)(0.050 s) 1.25 kg m/s x x J F t t = = + = + Thus 2 1 1.25 kg m/s (0.160 kg)( 3.00 m/s) x x x p J p = + = + + + = 1.73 kg m/s + 2 2 1.73 kg m/s 10.8 kg m/s (to the right) 0.160 kg x x p v m = = = + (b) (negative since force is to left) 2 1 ( ) ( 12.0 N)(0.050 s) 0.600 kg m/s x x J F t t = = − = − 2 1 0.600 kg m/s (0.160 kg)( 3.00 m/s) 0.120 kg m/s x x x p J p = + = − + + = − 2 2 0.120 kg m/s 0.75 m/s (to the left) 0.160 kg x x p v m = = = − E VALUATE : In part (a) the impulse and initial momentum are in the same direction and x v increases. In part (b) the impulse and initial momentum are in opposite directions and the velocity decreases. 8.16.I DENTIFY : Apply conservation of momentum to the system of you and the ball. In part (a) both objects have the same final velocity. S ET U P : Let + x be in the direction the ball is traveling initially. (ball). (you). 0.400 kg A m = 70.0 kg B m = E XECUTE : (a) 1 2 x x P P = gives and . 2 (0.400 kg)(10.0 m/s) (0.400 kg 70.0 kg) v = + 2 0.0568 m/s v = (b) 1 2 x x P P = gives 2 (0.400 kg)(10.0 m/s) (0.400 kg)( 8.00 m/s) (70.0 kg) B v = + and . 2 0.103 m/s B v = E VALUATE : When the ball bounces off it has a greater change in momentum and you acquire a greater final speed.

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